Question: $\log_{7}49 = {?}$
If $\log_{b}x=y$ , then $b^y=x$ First, try to write $49$ , the number we are taking the logarithm of, as a power of $7$ , the base of the logarithm. $49$ can be expressed as $7\times7$ $49$ can be expressed as $7^2$ $7^2=49$, so $\log_{7}49=2$.